Discuss Cable Calculations?! in the UK Electrical Forum area at ElectriciansForums.net

E

Emily B

Looking at putting a lathe into a workshop.. 16A, 3 phase, a run of 19metres at approx 35 degrees.

Also, a 3 phase, arc welder which I was going to put into a 32A socket outlet. 7 metre run, again at 35 degrees.

Can someone help with how I would complete a cable calculation for either?! Completely confused with three phase systems!

Cable will be run on perforated tray in an empty (non insulated) ceiling void.
 
Looking at putting a lathe into a workshop.. 16A, 3 phase, a run of 19metres at approx 35 degrees.

Also, a 3 phase, arc welder which I was going to put into a 32A socket outlet. 7 metre run, again at 35 degrees.

Can someone help with how I would complete a cable calculation for either?! Completely confused with three phase systems!

Cable will be run on perforated tray in an empty (non insulated) ceiling void.

Hi Emily.
Is this a homework question?
Tell us what you think. Show us your calcs and we are always happy to help.
We don't like just telling people the answer, as they don't learn anything.
 
Not a homework question, been given it by my boss and told to 'figure it out'.. I can complete single phase.. but unsure of how to calculate the design current at the start for 3 phase? I know the rest thing with correction factors etc
 
. but unsure of how to calculate the design current at the start for 3 phase?

You don't need to calculate the current.... do you?
You have a 16A supply and a 32A supply. I presume the breaker for the arc welder will be 32A? You just say you are using a 32A socket outlet.
So all you need to do is select the correct size cables.
Or am I missing something?
 
Its been a while but here goes:

Ib = Power (W) / Root 3 x Uo (V) x p.f

Where:

Ib = Design Current
Uo = Nominal Voltage (400V)
p.f. = Power Factor (Usually motors)

Then you select Device rating, In

Then you calculate Iz by applying correction factors : Iz = In/Cf

Ib < In < Iz

This is then used to size cable for current carrying capacity. Using this cable information you calculate the expected voltage drop:

VD(V) = mV/A/m x Ib x l
1000

This is where you would use your design current, but if you used In instead it would give the maximum expected volt drop for that circuit with the selected cable type.

Please be kind if i'm wrong with what I have posted as I believe that everyday is a school day.
 
It equal to or greater than
In/ ca cc cg ci

Without overload
Ib/ ca cc cv ci

Iz=It ca cc cg ci

And basically any other correction factor such as underground soil resistivity depth and such
 
what's soil resistivity got to do with it. you don't send current through soil.(hopefully)
 
Cables buried in the ground – soil thermal resistivity (Cs)
The current-carrying capacities tabulated for cables in the ground are based upon a soil thermal resistivity of 2.5K.m/W and are intended to be applied to cables laid in and around buildings, i.e. disturbed soil.

In locations where the effective soil thermal resistivity is higher than 2.5K.m/W, an appropriate reduction in current-carrying capacity should be made. Rating factors for soil thermal resistivities other than 2.5K.m/W are given in Table 4B3
 
Cs is the soil thermal resistivity, this is a completely different thing to soil resistivity.
Resistivity (rho) is the electrical resistance per unit area of a material.
Thermal resistivity is the reciprocal of the thermal resistance of a material.

These two physical properties are very different things,
 

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